Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random f

1. If x1 were a two-digit integer, then it would be explicitly stated: x1 is a tw-digit integer where x is a tens digit and 1 is an units digit.
2. On the real test subscripts would be clearly visible, not to create confusion: \(x_1\), \(x_2\), \(x_3\), \(x_4\), \(x_5\), \(x_6\), and \(x_7\), ... Edited above.

(1) seven numbers could be all different and the remainders also could be different. For example, if we have 13, 14, 21, 28, 35, 42, 56 the range of the remainders is 6 (6, 0, 0, 0, 0, 0, 0) the sum is also 6. However if we have 10, 11, 12, 13, 14, 15, 16 the range is still 6 but the sum is greater. The statement is not sufficient.

(2) if we have 10, 11, 12, 13, 14, 15, 16 the remainders will be 3, 4, 5, 6, 0, 1, 2 it is easy to see that there is a pattern of remainders, since the numbers are consecutive their remainders will always have a pattern from 0 to 6. Sufficient, thus the answer is B.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?

(1) The range of the remainders is 6.
(2) The seven integers are consecutive.

There are 7 variables, so we need 7 equations in order to solve the problem; only 2 equations are given, so there is high chance (E) will be the answer.
When we look at the conditions together, the remainders become 1,2,3,4,5,6,0 so the sum becomes 1+2+3+4+5+6=21, a unique answer. The condition is sufficient, so the answer seems to be (C), but this is a question with commonly made mistakes in 4(A) (Common Mistake Type 4 4(A)).
If we look at the conditions separately,
Condition 1 gives 1,2,3,4,5,6,7==> 1+2+3+4+5+6+0=21, but this also works for
7,14,21,35,42,48,49==>0+0+0+0+0+6+0=6, so this condition is not unique, and insufficient by itself.
Condition 2, on the other hand, the sum of the remainders becomes 1+2+3+4+5+6+0=21, so this condition is unique and therefore sufficient. So the answer becomes (B).

This type of question is a common type in today GMAT math

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Recall that the remainders vary from 0 to (n-1) when the divisor is n.

We pick 7 random integers and divide them by 7 so remainder in each case will be in the range 0 to 6 inclusive both.

(1) The range of the remainders is 6.
This tells us that one of the remainders was 0 and one was 6. It doesn't tell what the other remainders were. They could be 0, 3, 3, 3, 4, 4, 6 or 0, 1, 2, 3, 4, 5, 6 etc. Sum of remainders in different cases could be different. Not sufficient.

(2) The seven integers are consecutive.
If the integers chosen are consecutive, the remainders received will be 0, 1, 2, 3, 4, 5, 6.
Why?

Check this post: https://anaprep.com/number-properties-a ... -integers/

When we pick 7 consecutive integers, one of them will be a multiple of 7 and the others will be of the form (7a + 1), (7a + 2), (7a + 3), (7a + 4), (7a + 5) and (7a + 6). Hence the remainders obtained will be 0, 1, 2, 3,4 ,5 ,6 which have a defined sum of 21.
Sufficient alone.

Answer (B)

Check this video on Division and Remainders: https://youtu.be/A5abKfUBFSc

its a good question, I would go with B..

1) range tells me that maybe i got 34/7 etc...maybe the rest all multiples of 7 who knows..

2) tells us we got consecutive us , we got a series that repeats itself every 7 numbers..so thats suff

My answer will be B. Here is the explanation.

Clue 1 - Range is the difference between the largest and the smallest numbers. Possible remainders when a number is divided by 7 is 0 thru 6. SO range is 6. This can be derived from the question itself. So Clue 1 is the same derivation made is explicit. With this we cannot say what numebers are choosen and what would be ther remainders and in turn the sum. S0 Clue 1 is INSUFFICIENT.

Clue 2 - It says that the numbers are consecutive. This clearly tells that one number is divisible by 7. So remainder is 0. And the rest of the 6 numbers should leave the remainders from 1 thru 6. From this we can get the sum of the remainders. So Clue 2 is sufficient.

Any OA?

first step:
xi = 7*ki+ri where i=1 to 7.
sum of ri=?
(1) max-min (r1,....,r6)=6
(2) xi is such that x(i+1) = xi+1
So,
x1 = 7k1+r1
x2 = x1+1=7k1+r1+1
x3 = x2+1 = 7k1+r1+2
x7 = 7k1+r1+6.
sum of remainders = 7r1+1+2+3+4+5+6 = 7r1+21
one remainder is zero. so r1 can be solved for. hence B.

I didnt get why (1) is wrong.
I think all 7 integers are different. that is why if the range is 6, then there are consecutive integers that match the criteria. how is it possible to have the following, if 7 integers are different and the range of the set is 6? could you please write down any set that match this criteria?

0,1,2,3,4,5,6 -- range (6-0)=6
0,0,0,0,0,0,6 -- range (6-0)=6
0,6,6,6,6,6,6 -- range (6-0)=6

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Excellent Question
Here is what i did in this one =>
We need the sum of remainders
Statement 1=>
Clearly not sufficient.
It just tells us that there will be attest one number leaving a remainder 6 ad one number that is leaving a remainder zero.

Hence not sufficient

Statement 2-->
7 consecutive integers.
Forget the boundary 10,110 => If we pick any 7 consecutive integers =>WE will get all the remainders-> {0,1,2,3,4,5,6}Hence sufficient

Hence B

St 1 doesn't tell us anything that isn't already contained in St 2 because

11,12,13,14,15,16,16 will yield the same sum as

20,21,22,23,24,25,26


B

Small query, it says " x1 x2 x3 ... x7 " , Could not have one assumed that x is the tens digit and hence has the same tens digit: making A sufficient?